Finkelstein Test 3- Physics

8/12/2019 Finkelstein Test 3- Physics

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208 Formulae Sheet

Coulomb’s law: . Electric field created by a charge q: 204 r

q E

πε

=r

22F 1

04

1

r

qq

πε

=r

Permittivity of free space: 229229

0

/109/1094

1C m N coulombmeter Newton ⋅⋅=⋅×=

πε

Gauss’s law (electric flux through a closed surface): .Surface area of a sphere of radius R is 24 RS π =

A jump of the electric field over a charged surface:0

| || | E

σ δ

ε =

r

Electric potential of a point charge q: ( ) ( )r

qV r V 04πε

=∞−r . Unit: 1volt=J/C

ε0 = 8.85×10 -12 C/(Vm)

Definition of the electric potential difference:

Conservation of energy for a charge Q: const r QV K =+ )( .Energy of an electron in electric potential=1volt (electron volt): 1eV=1.6×10 -19 J(J=1Joule).

Capacitance: C=Q/V ; Parallel-plate capacitor: ε = K ε0

Spherical capacitor:1 1 1 1

4 inner outer C R Rπε

⎛ ⎞= −⎜ ⎟⎝ ⎠

; 4 a b

b a

R RC

R Rπε =

−

Unit:

ε0 = 8.85×10 -12 F/m

Capacitors in parallel: Capacitors in series:

Capacitor as energy storage

2 2

2 2Q CV

U C = = u=energy density

2 ( )( ) 2

E r u r ε =

rr

+++++++++++++++++++++++++++++++++++++++++++++++++

Definition of currentdt dQ

t Q

t I t

=ΔΔ=

→Δ 0lim)( . Unit: 1A=1ampere= C/s.

Current I= qnvS ( q=charge, n=density, v=velocity, S the cross-section area)

Ohm’s law: V/I=R; V=RI; V/R=I Unit: 1 Ω=V/A=Vs/C

Resistor with a constant cross section:sec ' Length L

Rcross tion s area S

ρ ρ = =− .

Resistivity ρ is measured in [Ωm].

∫ ⋅ ed d E ε

=S

enclosQS

0

rr∫ =⋅

S

enclosed QS d E 0ε

rr

0

(when an axes is directed from left to right !) E E σ

ε rhs lhs− =

2

1

2 1[ ( ) ( )]r

r

Edr V r V r = − −∫r

r

r r r r

d A

V Q

C ε = =Δ

1 1 1 /F farad coulomb volt ≡ =12 61 10 1 10 pF F F μ − −= = F

3...

1111

321

+++=C C C C tot

...21 + ++= C C C tot C



Resistors in series 321 R R R Rtot ++= +… Resistors in parallel ...1111

321

+++= R R R Rtot

Similarity between resistance and capacitance: 1/ 1/ R C ρ ε ⇔ ⇔

Power output (energy loss rate): . Unit: [J/s]2 2

/P IV RI V R= = =Discharging capacitor: ( ) exp( / )initialq t Q t RC = − ( )

/ q t

I dq dt RC

= = − ;

negative I implies that the charge flows out from the plate, i.e., it is dischargingCharging capacitor ( for the case when q(t=0)=0 ):

( ) [1 exp( / )] finalq t Q t RC = − − ( ) exp( / ) finalQ I t t

RC = − RC

Kirchhoff’s rules: sum of the directed currents in each of the junctions is zero;sum of the voltage drops and rises along each of the closed loops is zero.

++++++++++++++++++++++++++++++++++++++++++++++

Force acting on a charge q moving in the magnetic field

Force acting on an element dl of a current-carrying conductor:

Cyclotron frequency:

Dipoles. Electric dipole moment of a pair charges separated by : ;d ur

Magnetic dipole moment of a small area surrounded by a current I :

Torque [Nm]: p E τ = ×r urr u

; Bτ μ = ×r ur ur

.Energy of a dipole in a field: U p E = − ⋅

ur ur; U Bμ = − ⋅

ur ur

Magnetic field created by a moving charge q (Biot-Savart law):

Magnetic field created by an element dl carrying current I :

Units for magnetic field

Permeability

Magnetic field created by a straight wire carrying current I :

Steady-state version of Ampere’s law (current enclosed by a path):

Magnetic field created by a solenoid: 0 B nI μ = , n=N/l is number of turns per unit length.Faraday’s law (the EMF induced in a closed loop as response to a change of magnetic fluxthrough the loop):

Ampere’s law (including “displacement” current created by varying in time electricfields):

03

( )4

q v r B

r μ π

×=r rr

BvqF r rr ×=

Bl Id F rr

×=

2 2qB

f m

ω π π

= =

p qd =ur ur

±q

( ) I d S μ =ur ur

03

( )4

I d l r dB

r μ π

×=r rr

m A N smC N teslaT ⋅=⋅= /1//1)(1

7 7 2 20 4 10 / 4 10 / 4 10 /T m A N s C N Aμ π π π

− −

= × ⋅ = × ⋅ = ×7 2−

0 I | |2

Br

μ =π

∫contour

0 enclosed Bdl I μ =ur r

∫ Φ−=⋅dt

d r d E Brr

0 0( ) E c

d Bd r I

dtμ ε

Φ= +ur r

∫



Maxwell’s equations: two Gauss’s laws + Faraday’s and Ampere’s laws

Mutual Inductance: 22 2 2 2 2

B B

d 1 1 Emf N N M I

dt

Φ= − Φ = B

M I Φ

=21 12 M M = 11 1 1 1 12

B B

d 2 Emf N N M I

dt Φ= − Φ =

Mutual Inductance: 0 1 2mutual overlap overlap M n n l S μ =

Units for flux (weber) and EMF:2

2

[ ] 1 1 / 1 / 1

1 1 1 1 / B flux T m N m s C J s C V s

T m Wb V Wb s

Φ = ⋅ = ⋅ ⋅ = ⋅ = ⋅

⋅ = =

Units of the mutual inductance (henry):1 1 21 /1 1 /1 1 1 /henry H Wb A V s A s J A= = = ⋅ = Ω ⋅ =

Bd d

Emf N Ldt dt

I Φ= − = −Inductance (self-inductance):

Another units for permeability: 70 4 10 / H mμ π −= ×

Inductance of a toroidal solenoid:2

0 2 B N N

L A I r

μ π

Φ= = × rea

Current growth in an R-L circuit: (1 exp( / )) Emf

I Rt L R

= − −

Decay of current in an R-L circuit: ( 0)exp( / ) I I t Rt L= = −

Magnetic field energy:2 2( ) ( / )

( )2 2

LI t dQ dt U t L= =

Density of magnetic field energy2

02 Bu

μ =B

Oscillations in a L-C circuit:2

2

10

d qq

dt LC + = , ;2 1/ LC ω =

2 221 ( )( )

2 2 2( ) sin( )

q t Q LI t const

C C I t I t ω ϕ

+ = =

= +

++++++++++++++++++++++++++++++++++++++++++++++Waves (frequency, wave vector, speed): 2π 2π λ

ω = k = v = = ω /k ω = vk T λ T

Wave propagating along x: /( ; ) cos( )right left y x t A kx t phaseω ϕ ϕ = + =m

Wave equation: 2 22

2 2

( , ) ( , ) y t x y t xv

x t

∂ ∂=∂ ∂

Set of wave equations in electromagnetism:

Speed of light in vacuum and medium; index of refraction n:2 8 2

12 2 2 7 20 0

1 1(3 10 / )(8.85 10 / ) (4 10 / )c mC Nm N Aε μ π − −= = ≈ ×× ⋅ × s

0 0

( , )( , ) y z E t x B t x

x t ε μ

∂∂− =∂ ∂

2 2

0 02 2

( , ) ( , y y E t x E t x

x t ε μ

∂ ∂=

∂ ∂( , ) ( , ) y z

E t x B t x x t

∂ ∂= −∂ ∂

)



2 1

/ /magnv n c v KK K v c nεμ

= = = ≈ =

Relation between the amplitudes of the electric and magnetic fields in electromagnetic

fields: E=cB.

Radiation power: P=IA Intensity of radiation far away from the source:

2/(4 ) I P r π = Density of energy: ; average density of energy2

0 ( , )u E x t ε = 2 20 0( , ) / 2u E x t E ε ε = =

Poynting vector S , intensity I :

Radiation pressure: /rad P I cα = ; for totally reflecting mirror α =2; for black body α =1. 0 02

E B E S P S A I S

Bμ μ ×= = ⋅ =

u

=r ur

ur ur ur

+++++++++++++++++++++++++++++++++++++++++++++++++++Angle of reflection:

incident reflected θ θ =

sin sinincident incident refracted refracted n nθ θ =Snell’s law:

Angle of total internal reflection: sin refracted critical

incident

n

nθ =

Polarizing by a linear filter along the direction

Malus’s law (consequence of the relation above):

Polarizations:

Bruster’s angle: refracted polar

incident

ntg

nθ =

Huygens’s and Fermat’s principles.

Images; lateral magnification:

Concave spherical mirror → ), focal length:

Convex spherical mirror → ( :

Spherical refractive image

Thin lenses (converging lens, f>0; diverging lens, f<0):1 1 1 '

's

ms s f s

+ = = −

Lens maker’s equation:

Double convex/concave lenses:

; ;2 2

: .

y z y z y z y z

y z

circular E E elliptical E E

linear

π π ϕ ϕ ϕ ϕ

ϕ ϕ

= = ≠ =

=

m m

' ' y sm y s= = − 1 1 1

| | /'

f Rs s f

+ = = 2

1 1 1| | | | / 2

' | | f R

s s f + = =

−' '

'

a b b a a

b

n n n n y n sm

R y n ss s−+ = = = −

1 1 1( 1)

| | | | | |n

f R R

⎛ ⎞= − +⎜ ⎟⎝ ⎠

( ): filter incident incident n n E → ⋅n E

r ur r r ur

2max cos I I φ =

1 2

1 1( 1)n

1

f R R

⎛ ⎞= − −⎜ ⎟

⎝ ⎠



++++++++++++++++++++++++++++++++++++++++++++++Integrals:

1

11 +

∫ += N N

x N dx x 1

0 11 +

∫ += N

R N

R N dx x

1

11

11−∫ +−

= N N x N

dx x

∫∞

−−=

R N N R N

dx x 1

11

11

∫ = xdx x

ln1

∫ =b

a

abdx x

)/ln(1

( )3/ 2 2 20 2 2

1 1a a x dy x x a x y = ++∫

0 0 00exp( / ) (1 exp( / )

t d t τ τ τ τ τ − = − −∫

2 2 1cos ( ) sin ( )

2cos( ) sin( ) 0

cos / 6 3 / 2 sin / 6 1/ 2

cos / 3 1/ 2 sin / 3 3 / 2

t kx t kx

t kx t kx

ω ω

ω ω

π π

π π

− = − =

− ⋅ − =

= == =

Averaging



Physics 208, Spring 2010 – Exam #3 version A

Name (Last, First): ___________________________________

ID #: ___________________

Section #: __________________

Yang 501, Nathan 502, Wei 1pm 503, Matt 504, Wei 5pm 505 ___________________________________________________________________________

· You have 70 minutes to complete the exam.

· Formulae are provided on a separate sheet. You may NOT use any other formula sheet.

· You may use only a simple calculator: one without memory, or with a memory demonstrated to be cleared.

· When calculating numerical values, be sure to keep track of units. Results must include proper units.

· Be alert to the number of significant figures in the information given. Results must have the correct number of significant figures, =3.14.

· If you are unable to solve a problem whose solution is needed in another problem, then assign a symbol for the solution of the first problemand use that symbol in solving the second problem.

· If you need additional space to answer a problem, use the back of the sheet it is written on.

· Show your work. Without supporting work, the answer alone is worth nothing.

· Mark your answers clearly by drawing boxes around them.

· Please write clearly. You may gain marks for a partially correct calculation if your work can be deciphered. __________________________________________________________________________________________________________________

1 (25) 2 (35) 3 (15) 4 (25) 5 (20) 2-Bonus(10)

3-Bonus(15)

5-Bonus(10)

Total(120+35



Physics 208, sections 501-505 Alexander Finkel’steinApril 20, 2010

1. Cyclotron motion + Ampere’s law (25 grades)

Two parallel wires carrying parallel-directed currents, I=1A, are 1 m apart from each other(L=1m), see Figure. We are interested in the point p located on x-axes at a distancea= 3 / 2 m from the origin. In other words, the point p is 1m away from each of the wires.(The rectangular triangle with angle 30degree.)

a) What is the value of the magnetic field B in the point p ? How is it directed? b) What is cyclotron frequency f of an electron in this point?c) What is the direction (clockwise or anti-clockwise) of rotation of the electron if one lookson it from above. (caution: the electron charge is negative).The charge of the electron , and its mass191.6 10e −= × C g 319.1 10m k −= × .

Answer: B=f =

direction:

y

x p z

I

I

L

a

29229

0

/109/1094

1 C m N coulombmeter Newton ⋅⋅=⋅×=πε

eeeee

r

qqF

221

041

πε =

r



2. Rail-gun (a slide wire motor). (35 grades) A battery supplies current I into a circuit which includes a sliding rod and resistance(the rod is in contact with two metal rails, see Figure). The system is placed in a

uniformmagnetic field B=20T. The rails are at a distance L=1m apart. The resistance of thecircuit (rails, rod, plus internal resistance of the battery) is R=20 Ω. At a given moment, the current I induces a force F which acts on the rod to the right,F=80N. At the same moment the velocity v of rod, which is moving to the right, is equalv=10m/s.

he rod at

e motor’s back EMF generated by

power consumed by

?f) What power is supplied by the battery?

): what is maximal velocity which can be developed by the rail-gun with thisbattery.

=P diss = P batt =

Bonus: vmax=

a) What current I has to flow through t

x x x x

x x x x

x x x

x x xx

x x

x x

x x xx

x x x x

x x x

x x x x

the discussed moment; in what direction? b) What is ththe moving rod?c) What is the EMF of the battery?d) What is the mechanicalthe rod at the discussed moment?e) What power is dissipated by the resistor

Bonus (10

Answers: I= back-EMF= EMF battery

P rod =

B

x

x

x

FBattery



3.s

is R=2 Ω.speed 1A/s (1 ampere per second).

up? b) What current is flowing inside the coil?

nswer: M= I=

at is the totalcharge that will pass along the coil, R=2 Ω, as a result of this move?

Answer: Q=

Mutual Inductance. (15 grades) An ideal solenoid with length L=1m and a cross-sectional area S=20cm 2 has 1000 turns. It isurrounded by a metallic coil with one turn, see Figure. The coil is located in the middle ofthe solenoid. The resistance of the coil The current in the solenoid is changing witha a) What is the mutual inductance of this set

A

bonus (15): The current inside the solenoid is maintained to be 1A and is constant in time. The coil is moved to the far left away from the solenoid, see Figure. Wh



4. LC-circuit (25 grades)

steady current is equal 2A.e

a) heimal charge that the

b) losingaximally

charged.

max max

In the circuit the switch S 1 has been closed for long time. TheSuddenly, switch S 1 is opened and S 2 is closed at the sam

instant. Capacitor C=20 μF and an inductor L=80 μH.Assuming that at the initial moment the charge on tcapacitor is zero, find the maxcapacitor will receive, Q max .Find the time t max (counting from the moment of cswitch S 2) when the capacitor will be m

Answer: Q = t =



5. Self-inductance (inertia) (20 grades)

In the circuit shown in the Figure, ε = 30.0 V, R1 = 20.0 Ω , R2 = 15.0 Ω and L =.500 Η . Switch S is closed at t = 0. Just after the switch is closed, find

resistor R2 ?

tential difference, vab , across

) Which point a or b is at higher potential?

thened,

difference,

tential difference, vab , across

) Which point a or b is at higher potential?

int: sum of potential differences along any closed contour is equal zero.

0): after closing switch S (at t=0), the current initially grows linearly with time I=Jt .ind J.

nswers:

0 a) what is the potential difference, vcd , across

b) Which point c or d is at higher potential?c) What is the pothe inductor L?d The switch is left closed a long time and opened. Just after the switch is opene) what is the potentialvcd , across resistor R2 ?f) Which point c or d is at higher potential?g) What is the pothe inductor L?h H

Bonus (1F A

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Finkelstein Test 3- Physics